Page 521 - Advanced engineering mathematics

14.7 DFT Approximation of the Fourier Transform 501

Using these numbers, we can compute
127
πik/2
V k e
k=0
=127 + (−1 + 40.735i)e πi/2 + (−1 + 20.355i)e πi
+ (−1 + 13.557i)e 3πi/2 + (−1 + 10.153i)e 2πi
+ (−1 + 8.1078i)e 5πi/2 + (−1 + 6.7415i)e 3πi
+ (−1 + 5.7631i)e 7πi/2 + (−1 + 5.0273i)e 4πi
+ (−1 + 4.4532i)e 9πi/2 + (−1 + 3.9922i)e 5πi

+ (−1 − 3.9922i)e 118πi/2 + (−1 − 4.4532i)e 1119πi/2
+ (−1 − 5.0273i)e 120πi/2 + (−1 − 5.7631i)e 121π/2
+ (−1 − 6.7415i)e 122πi/2 + (−1 − 8.1078i)e 123πi/2
+ (−1 − 10.153i)e 124πi/2 + (−1 − 13.557i)e 125πi/2
+ (−1 − 20.355i)e 126πi/2 + (−1 − 30.735i)e 127πi/2
=61.04832.

Then
127
1 πijk/64
V k e = 0.47694.
128
k=0
This gives the 128 point DFT approximation of 0.47694 to the sampled partial sum S 10 (1/2),
which was computed to be 0.45847. The difference is 0.0185.
The actual sum of the Fourier series at t = 1/2is f (1/2) = 0.50000. In practice we would
achieve greater accuracy by choosing N larger, allowing larger M.

SECTION 14.6 PROBLEMS

In each of Problems 1 through 6, a function is given 1. f (t) = 1 + t for 0 ≤ t < 2, p = 2, t 0 = 1/8
2
having period p. Compute the complex Fourier series 2. f (t) = t for 0 ≤ t < 1, p = 1, t 0 = 1/2
of the function and then the 10th partial sum of 3. f (t) = cos(t) for 0 ≤ t < 2, p = 2, t 0 = 1/8
this series at the indicated t 0 . Then, using N = 128, 4. f (t) = e −t for 0 ≤ t < 4, p = 4, t 0 = 1/4
3
compute a DFT approximation to this partial sum 5. f (t) = t for 0 ≤ t < 1, p = 1, t 0 = 1/4
at t 0 . 6. f (t) = t sin(t) for 0 ≤ t < 4, p = 4, t 0 = 1/8.

14.7 DFT Approximation of the Fourier Transform

Under certain conditions the DFT can be used to approximate the Fourier transform of a function.
ˆ
To begin, suppose f (ω) can be approximated to within some acceptable tolerance by an integral
over a finite interval
∞
2π L
ˆ
f (ω) = f (ξ)e −iωξ dξ ≈ f (ξ)e −iωξ dξ.
−∞ 0

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October 14, 2010 16:43 THM/NEIL Page-501 27410_14_ch14_p465-504